1. Nov 15, 2005

abot

in the attachment there is a problem related to techniques of integration, i fully understand everything until near the end of the problem they answer says

(1/6)o - (1/12)sin2o + C = (1/6)o - (1/6)sino coso + C

and then they convert the sinocoso to sec-1

this part is really confusing...can you help me plz...

thank you

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• math_problem.doc
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2. Nov 16, 2005

benorin

Recall that $$\sin(2o)=2\sin (o)\cos (o)$$, this is the first one.

Note that, from the triangle, we have $$\sec (\theta ) = \frac{x}{3} \Rightarrow \theta = \sec ^{-1} \left( \frac{x}{3}\right)$$
and, also from the triangle, we have $$\cos (\theta ) = \frac{3}{x}$$

and $$\sin (\theta ) = \mbox{ what? }$$, and the required result follows.